What is the vertices y=3x^2-30x+77 ?

The vertices y=3x^2-30x+77 is Minimum (5,2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=3x^2-30x+77

vertices

y = 3 x^{2} - 30 x + 77

Minimum (5, 2)

Solution steps

y = 3 x^{2} - 30 x + 77

The parabola parameters are:

a = 3, b = - 30, c = 77

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{( - 30 )}{2 \cdot 3}

Simplify

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(5, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (5, 2)

y = x^{2} - 4

y = x^{2} + 1

4 x^{2} - 24 x + 31

y = x^{2} + x

y = 9 x^{2}

What is the vertices y=3x^2-30x+77 ?
The vertices y=3x^2-30x+77 is Minimum (5,2)